JEE Advanced Mock Test - 5 (Paper I) - JEE MCQ

Let the spring constant of spring-1 be k₁ and its amplitude be A₁.

Similarly the spring constant of spring-2 be k₂ and its amplitude be A₂.

Net Amplitde A = A₁ + A₂ .........(1)

When displacement is maximum, force acting on each spring is shown in figure.

Internal restoring forces acting in spring-2 is k₂A₂. Restoring force acting in spring-2 pulls spring-1.

Since restoring force acting in spring-1 is k1A1, at the time of maximum displacement,

k₁A₁ = k₂A₂ .............(2)

Using eqn.(1), we write eqn.(2) as, k₁A₁ = k₂ (A - A₁)

Hence from above eqn., we get, A₁ = [k₂/(k₁ + k₂)] A

The diagrammatic representation of the given problem is shown in figure. The expression of fundamental frequency is

In air,

T = mg = (Vρ)g

… (i)

When the object is half immersed in water,

T' = mg - up thrust = Vρg -

The new fundamental frequency is

Or

Let tension in the wire is T, then using free body diagrams for both the blocks:

1) For mass m,

ma = T − mg

2) For mass 2m,

2ma = 2mg − T

Thus,

Probability of survival,

For one mean life, t = 1/λ

P(survival) = 1/e

At A,

At B:

At C :

At D :

From figure r_B < />_Dhence N_B > N_D

Hence N_B is greatest

r_C < />_A

N_C < />_A

Hence N_C is least

At A C ; N_A < />

N_C < />

At B D ; N_B > mg

N_D > mg

Pressure due to liquid column is equal to hρg only if fluid is at rest

Hence, (A) is not correct and (B) is correct.

At same horizontal level pressure decreases with velocity hence, (C) is correct

P_A = P_E = Atmospheric pressure

From graph ⇒ T = 0.2 sec And amplitude of standing

Wave is 2A = 4 cm

Equation of the standing wave

speed = λ/T = 2 m/sec

= 20π cm/sec

Let the moment of inertia of the bigger disc be  

Acceleration of A down the plane,

a_A = g sin 45° -μ_A g cos 45^o

=

= 4√2m/s²

Similarly acceleration of B down the plane, a_B =g (sin 45°- μ_B cos 45°)

=

The front face of A and B will come in a line when,

s_a = s_B + √2

Solving this equation, we get t = 2 s

Additional information,

So, both the blocks will come in a line after A has travelled a distance 8√2 m down the plane.

As we have Q = mL

Hence

If R is the thermal resistance of the given material , then heat flow is

Where dT is the temperature difference across the two points

From 1 and 2

When rods are connected in parallel, the thermal resistance is R/2

When rods are connected in series, the thermal resistance is 2R

Let us say, PO = OQ = XR = a

Applying

Substituting the values with sign, we have

(Distance are measured from O and are taken as positive in the direction of ray of light.)

a = 5R

and a = XR

Hence X = 5

(A) When a wire is connected across a charged capacitor, the capacitor gets discharged immediately.
(B) When a wire is moved perpendicular to its length with a constant velocity in a uniform magnetic field perpendicular to the plane of motion, free electrons will experience force due to the magnetic field and accumulate at the one end and the other positive charge will accumulate at the other end which will create a constant potential difference.
Therefore,
(A) - (u)
(B) - (r)

Let x be the small displacement of the centre of the mass of the sphere and θ be the angular displacement of centre of mass w.r.t point of contact, then
X=Rθ
Hence,

Total energy of the system = Linear kinetic energy + Rotational kinetic energy + Potential energy of the spring

As the total energy remains constant,

Hence, angular frequency

Equivalent spring constant is k_eq = ⁵⁄₇k
Total energy of the SHM ¹⁄₂kA².
At x = ^A⁄₂, total energy

Let x be the small displacement of the block and θ be the angular displacement of pulley, then
X=Rθ
Hence,

Total energy of the system = Linear kinetic energy of the bock + Rotational kinetic energy of the pulley + Potential energy of the spring

As the total energy remains constant,

Hence, angular frequency

Equivalent spring constant is k_eq = 2/3k
Total energy of the SHM ¹⁄₂kA².
At x = ^A⁄₂, total energy

Let x be the small displacement of centre of mass of the hollow sphere and be the angular displacement with respect to the point of contact. Then,
x= Rθ

Displacement of the point P = 2x
Hence, the elongation in the spring is 2x.
Velocity of the centre of mass of the hollow sphere

Total energy of the system = Linear kinetic energy of the sphere + Rotational kinetic energy of the sphere + Potential energy of the spring

As the total energy remains constant,

Hence, angular frequency

 

Equivalent spring constant, k_eq = (12/5)k
Total energy of the SHM = 1/2kA²
At x = A/2, total energy is given by the formula:
Total energy = Potential energy + Kinetic energy

ΔG = -nFE = -(2 x 96500 x 0.059) = -11387 J = -11.387 kJ mol^-1

Step (ii) involves the substitution of the -OH group to form acid chloride.

Step (iii) involves acylation of the benzene ring, in the presence of a Lewis acid as a catalyst, following the Friedel Craft's mechanism.

Units of R

(i) In L atm ⇒ 0.082 L atm mol⁻¹K⁻¹

(ii) In C.G.S. system ⇒ 8.314×107erg mol⁻¹K⁻¹

(iii) In M.K.S. system ⇒ 8.314 J mol⁻¹K⁻¹

(iv) In calories ⇒ 1.987 cal mol⁻¹K⁻¹

1. Guanidine is a stronger base than pyridine. The increased basicity can be explained by drawing the resonance structures of the protonated guanidine.

2. Dimethylamine is more basic than trimethylamine in aqueous medium. This is due to the fact that steric hindrance created by alkyl groups affect the solvation so badly that the tertiary amines are usually the least basic among the three classes. Due to the presence of 3 alkyl groups, there is a lot of steric hindrance and therefore, it is difficult for nitrogen to form H-bonds with the hydrogen atom of water for proper solvation.

3. Ortho one is more basic due to the inductive effect of the methyl group.

4. 2,4,6-Trinitro-N,N-dimethyl aniline is a stronger base than 2,4,6-Trinitroaniline because the steric repulsion of -N(CH₃)₂ goes out of the plane of the ring.

Ferric (III) chloride test (violet colour):

Molisch test: It is a sensitive chemical test, for the presence of carbohydrates, based on the dehydration of the carbohydrate by sulfuric acid or hydrochloric acid to produce an aldehyde, which condenses with two molecules of a phenol.

EDTA: 6; two N and four O
Ethylenediammine: 2; two N

Diethylenetriammine: 3; three N
Triethylenetetramine: 4; four N

Aldehyde and ketone, on addition with HCN, give cyanohydrin.

Methanal does not have any α-Hydrogen atom.
In the presence of a strong base, it undergoes disproportionation to form methanol and a salt of methanoic acid.

A.

From equations (1) and (2), we have
e = 1/√3
B.
Equation of the director circle would be:
x² + y² = 9

C.

D.
Here, we have
x + y – 2 = 0 …… (1)
x – y = 0 …… (2)
Solving equations (1) and (2), we have
(x, y) = (1, 1)

A.

B.

C.

Here, the equation is of the form of    so the solution becomes,

D.

Here, the equation is of the form of    so the solution becomes,
e^x + x² y = C y
where C is an arbitrary constant.

Here, we have to find the general solution for each differential equation or we can check for individual solution.

Integrate both sides.

Compare the above result with expression in column IV.
Therefore, y = x³e^2x and C = 0

Integrate both sides.

Compare the above result with expression (IV).


From the above result,


Compare the above result with expression (II).


Integrate with respect to x.

Compare the above result with expression (I).

Hence, (ii), (III), (A) is the correct combination.